# Transitivity of $R$ when it is a relation such that $R^2 = R$ and Vice-Versa.

My question is about Set and Relation Theory.

$$R$$ is a relation on a set $$S$$.

1) Show that if $$R^2 = R$$ then $$R$$ is transitive.

2) Show that if $$R$$ is transitive and "Reflexive or Symmetric" Then $$R=R^2$$. (It means that for a transitive and reflexive relation we have $$R = R^2$$ and for a transitive and Symmetric relation we have $$R=R^2$$)

For 1, My proof is: Let $$aRb$$ and $$bRc$$ then by definition of R^2 we have $$a R^2 b$$ and because $$R^2 = R$$, we have $$a R c$$. so $$R$$ is transitive.

But I don't know what to do for 2.

Is my answer to "1)" correct? How to approach 2?

• Can you prove that if $R$ is transitive, then $R^2\subseteq R$? What are you missing to be able to prove the other inclusion? How would “reflexive” help? How would “symmetric” help? – Arturo Magidin Jun 1 '19 at 18:33
• @ArturoMagidin yes. I can prove that. but we will also need to show that $R \subseteq R^2$. Can i get this from "reflexive" or "symmetric"? – amir na Jun 1 '19 at 18:35
• @ArturoMagidin Actually I need to prove that if $aRb$ then there exist a c such that $aRc$ and $cRb$? So I think when it is reflexive that c is $a$ yes? – amir na Jun 1 '19 at 18:37
• @ArturoMagidin also, If we exclude the empty elements that are isolated, we can say that for the rest, if relation is symmetric and transitive, it is also reflexive. yes? – amir na Jun 1 '19 at 18:41
• That was my question: how can you use “reflexive” to get $R\subseteq R^2$? Yes, you need to show that if $aRb$, then there exists $c$ such that $aRc$ and $cRb$. Would “reflexive” help you there? How about “symmetric”? – Arturo Magidin Jun 1 '19 at 18:51

## 1 Answer

Your answer to the first question is correct.

Now, suppose that $$R$$ is transitive and reflexive. You want to prove that $$R=R^2$$. If $$a\mathrel Rb$$, then, since $$b\mathrel Rb$$ and $$R$$ is transitive, $$a\mathrel{R^2}b$$. So, $$R\subset R^2$$. And if $$a\mathrel{R^2}b$$, then there is a $$c$$ such that $$a\mathrel Rc$$ and $$c\mathrel b$$. So, since $$R$$ is transitive, $$a\mathrel Rb$$. This proves that $$R^2\subset R$$. So, $$R^2=R$$.

Finally, suppose that $$R$$ is transitive and symmetric. Again, you want to prove that $$R=R^2$$. If $$a\mathrel Rb$$, then, since $$R$$ is symmetric, $$b\mathrel Ra$$. And, since $$b\mathrel Ra$$, $$a\mathrel Rb$$ and $$R$$ is transitive, $$b\mathrel Rb$$. Now, since $$a\mathrel Rb$$, and $$b\mathrel Rb$$, $$a\mathrel{R^2}b$$. So, $$R\subset R^2$$ and you can prove that $$R^2\subset R$$ as above.

• I am now sure that my answer to the second question in comments is also right. Thank you very much. – amir na Jun 1 '19 at 18:43